find a correlation

I have a Matrix

 X=[0.231914928  3.126057882  -1.752476846
-0.779092587  2.143243132  -1.944363312
-1.744892449  1.206497824  -2.267829067
-0.276817947  1.774687601  -1.768924258
-0.367233254  1.697905199  -1.508506912
-0.367233254  1.697905199  -1.508506912
-1.378240769  0.814907572  -1.700393377
-2.389248284  -0.060411815  -1.892279842
-1.333033116  0.860977013  -1.831972668
0.135041386  1.40613207  -1.333067858]

 Y=[0.253549664
-0.231692981
0.768395971
2.988670669
-0.038625616
-0.038625616
-0.525155376
-1.011685136
0.961463336
3.181738034]

At first I want to calculate the correlation coefficient between all X columns which can be done like this

[R]=corrcoef(X)

then I want to see which pair of columns has the highest correlation together for example column 1 with 2 ? 1 with 3? 2 with 3?

then the one that has correlation more than 0.5 lets say for example columns 1 and 2 , then check their correlation with y and say which one is more correlated

[Daniel Shub]

Mohammad, in general, I think your questions are great questions for Answers, but often I feel that I do not understand your questions. I think it would be easier for those of us who chose to answer questions, if you could spend a little more time composing your questions.

[Niki]

Thanks Daniel for your comment, For sure, I will do my best , Thanks

Answer 1

[Grzegorz Knor]

For highest correlation:

maxR = max(max(triu(R,1))) % highest correlation
[row,col] = find(R==maxR,1,'first')

The greatest correlation occurs between the columns row an col.

Second question:

[r w] = max([corr(X(:,row),Y) corr(X(:,col),Y)])

w equal to 1 means that X(:,row) has a higher correlation than X(:,col). w equal to 2 mean that X(:,col) has a higher correlation than X(:,row).

[Niki]

there is one problem

at first it only works for the highest correlation , if there are several then will not work, so lets say find the pair correlation in X higher than 0.5 , and then

regarding to the second question

R =

1.0000 0.8171 0.6448

0.8171 1.0000 0.1545

0.6448 0.1545 1.0000

means the highest correlation is between columns 1 and 2

but the command

[r w] = max([corr(X(:,row),Y) corr(X(:,col),Y)])

only calculate the correlation between column 2 and Y

and what about column 1 and Y?

[Grzegorz Knor]

The command

[r w] = max([corr(X(:,row),Y) corr(X(:,col),Y)])

calculates the correlation between column 2 and Y and between 1 and Y. Look:

[corr(X(:,row),Y) corr(X(:,col),Y)]

[Grzegorz Knor]

For first question:

[row,col] = find(triu(R,1)>=0.5)

Then you can use function corr.

[Niki]

please take a look at the code that I used , and if you can solve the error

Thanks

[Niki]

if you put the command, I can accept your answer

[Niki]

corr(X(:,unique([row;col])),Y)

Answer 2

[the cyclist]

It wasn't perfectly clear to me if you wanted to find correlation with Y for all the columns that had r>0.5, or only the highest. This does all of them. Maybe you could tailor this to what you need.

r = corrcoef(X); % Correlation coefficiant

[i j] = find(r>0.5); % Indices of r > 0.5

indexToOffDiagonalElementsWithHighCorrelation = (i~=j); % Only use off-diagonal elements

XColumnsWithHighCorrelation = unique(i(indexToOffDiagonalElementsWithHighCorrelation))

for nx = 1:numel(XColumnsWithHighCorrelation)

      rxy{nx} = corrcoef(X(:,XColumnsWithHighCorrelation(nx)),Y);
      disp(rxy{nx})

end

[Niki]

The first problem when you perform

>>r = corrcoef(X)

then you will have like this

r =

1.0000 0.8171 0.6448

0.8171 1.0000 0.1545

0.6448 0.1545 1.0000

But I would like to have like this

r =

0 0 0

0.8171 0 0

0.6448 0.1545 0

How can I do that ?

[Oleg Komarov]

use triu to zero out the upper diagonal.

[Niki]

Thanks Oleg, I did not know " Triu " :D

Andrei put a command for that, Thanks

Answer 3

[Andrei Bobrov]

[v id]= max(triu(corrcoef([X,Y]),1))

Variant last

R = triu(corrcoef([X,Y]),1) 
Rx = R(1:end-1,1:end-1) 
Rx05 = Rx.*(Rx>.5) 
[ix jx] = find(Rx05==max(Rx05(:))) 
cYX = R([ix,jx],4) 
[vXY xi]= max(cYX)

[Niki]

Andrei

when I use of your command

[v id]= max(triu(corrcoef([X,Y]),1))

v =

0 0.8171 0.6448 0.5133

what is the output?

first is 0

second is 0.8171 (between column 1 and 2)

third is 0.6448 (between column 1 and 3)

the last is 0.5133 (???)

[Andrei Bobrov]

last between X(:,1) and Y , because v(4) = 1

[Niki]

it is 1 because the highest was column 1 and column 2

therefore if i want to also have the X(:,2) and Y. then what should I do?

[Andrei Bobrov]

R = triu(corrcoef([X,Y]),1)

Rx = R(1:end-1,1:end-1)

Rx05 = Rx.*(Rx>.5)

[ix jx] = find(Rx05==max(Rx05(:)))

cYX = R([ix,jx],4)

[vXY xi]= max(cYX)

[Andrei Bobrov]

X(:,2) and Y -> R(2,4)

[Niki]

I think you can not reach to the answer with this command please check this out

for example we have

>> X=rand(10);

>> Y=rand(1,10);

then we perform

>>[R]=corrcoef(X);

then if you perform

>>[v id]= max(triu(corrcoef([X,Y]),1))

v is different with R , which I only can see one value similar, could you please tell me what is happening with this command ?

[Niki]

Andrei, I like your comment very much, Thanks